SOLUTION: Write the equation of the circle satisfying the center (3,4) and tangent to 2x

SOLUTION: Write the equation of the circle satisfying the center (3,4) and tangent to 2x - y + 5 = 0

Question 1119397: Write the equation of the circle satisfying the center (3,4) and tangent to 2x - y + 5 = 0
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Write the equation of the circle satisfying the center (3,4) and tangent to
2x - y + 5 = 0
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It's the line thru (3,4) perpendicular to the given line.
Can you do that?
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The slope of the given line is 2.
A line from a tangent point to the center is perpendicular to the tangent line.
Lines perpendicular to the given line have a slope that's the negative inverse:
Slope m = -1/2
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Use y-y1 = m*(x-x1)
y-4 = (-1/2)*(x-3) is an equation of the line asked for.
----
2y-8 = -x+3
x + 2y = 11
Equations of the same line, just manipulated and "simplified."
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The intersection of the 2 lines is the tangent point. The radius is the distance from the center to the tangent point.
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A shorter method:
The distance from a point (x,y) to a line ax + by + c = 0 is
|ax + by + c|/sqrt(a^2 + b^2)
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(3,4) to 2x - y + 5 = 0
d = |2*3 - 4 + 5|/sqrt(2^2 + 1^1) = 7/sqrt(5) = r
r^2 = 49/5
--> (x-3)^2 + (y-4)^2 = 49/5 is the circle

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